We solve this task in two different ways: (i) by an analytical approach (with the “pocket calculator”) and (ii) numerically by running the whole gamut of chemical thermodynamics. The results will differ considerably, and we will explain why.

Since the dissolved amount of gypsum equals the value of [SO4], we immediately get the result:

(5)

ΔGypsum

=

K1/2

(6)

ΔGypsum

=

(10-4.58)1/2 M

(7)

ΔGypsum

=

10-2.29 M = 5.12×10-3 M = 5.12 mM

In words: The amount of 5.12 mmol gypsum dissolves in 1 liter of pure water. [Please note that Eq.(5), and the final result, are independent of the initial inventory of gypsum.]

Numerical Approach with aqion

We start with pure water (button New) and switch to molar units (activate checkbox Mol). To open the mineral table click on Minerals, then enter the amount of 20 mmol/L gypsum.5

Run the calculation with Start, a first schematic overview appears. After click on next, the results are shown again in the output table. In the right column you find:

pH

=

7.07

Ca

=

15.6 mM

SO4

=

15.6 mM

According to this calculation, the amount of 15.4 mmol gypsum dissolves in 1 liter of pure water – it is three times more than in the calculation above!

Here, SO4 and Ca abbreviate the total concentrations:

(8)

[Ca]T

=

[Ca+2] + [CaSO4(aq)] + [CaHSO4+] + [CaOH+]

(9)

[SO4]T

=

[SO4-2] + [HSO4-] + [CaSO4(aq)] + [CaHSO4+]

The corresponding equilibrium speciation is listed in the table Ions:

Ca+2

10.5

mM

CaSO4(aq)

5.19

mM

CaHSO4+

3.2×10-6

mM

CaOH+

1.2×10-5

mM

SO4-2

10.5

mM

HSO4-

5.1×10-5

mM

In fact, there are only three major players: Ca+2, SO4-2 and CaSO4(aq). The last species is an aquoeous complex; it shoud not be confused with the mineral phase gypsum usually abbreviated with CaSO4(s).